On various (strong) rainbow connection numbers of graphs

Transcription

1 On various (strong) rainbow connection numbers of graphs Lin Chen 1, Xueliang Li 1, Henry Liu 2, Jinfeng Liu 1 1 Center for Combinatorics and LPMC Nankai University, Tianjin , China chenlin @126com, lxl@nankaieducn, ljinfeng709@163com 2 School of Mathematics and Statistics Central South University, Changsha , China henry-liu@csueducn 18 April 2017 Abstract An edge-coloured path is rainbow if all the edges have distinct colours For a connected graph G, the rainbow connection number rc(g) is the minimum number of colours in an edge-colouring of G such that, any two vertices are connected by a rainbow path Similarly, the strong rainbow connection number src(g) is the minimum number of colours in an edge-colouring of G such that, any two vertices are connected by a rainbow geodesic (ie, a path of shortest length) These two concepts of connectivity in graphs were introduced by Chartrand et al in 2008 Subsequently, vertex-coloured versions of both parameters, rvc(g) and srvc(g), and a total-coloured version of the rainbow connection number, trc(g), were introduced In this paper we introduce the strong total rainbow connection number strc(g), which is the version of the strong rainbow connection number using total-colourings Among our results, we will determine the strong total rainbow connection numbers of some special graphs We will also compare the six parameters, by considering how close and how far apart they can be from one another In particular, we will characterise all pairs of positive integers a and b such that, there exists a graph G with trc(g) = a and strc(g) = b, and similarly for the functions rvc and srvc Keywords: (Strong) rainbow connection; (strong) rainbow vertex-connection; (strong) total rainbow connection AMS Subject Classification (2010): 05C12, 05C15, 05C40 1 Introduction In this paper, all graphs under consideration are finite and simple For notation and terminology not defined here, we refer to [4] Corresponding author 1

2 In 2008, Chartrand et al [7] introduced the concept of rainbow connection of graphs An edge-coloured path is rainbow if all of its edges have distinct colours Let G be a non-trivial connected graph An edge-colouring of G is rainbow connected if any two vertices of G are connected by a rainbow path The minimum number of colours in a rainbow connected edge-colouring of G is the rainbow connection number of G, denoted by rc(g) The topic of rainbow connection is fairly interesting and numerous relevant papers have been published In addition, the concept of strong rainbow connection was introduced by the same authors For two vertices u and v of G, a u v geodesic is a u v path of length d(u, v), where d(u, v) is the distance between u and v An edge-colouring of G is strongly rainbow connected if for any two vertices u and v of G, there is a rainbow u v geodesic The minimum number of colours in a strongly rainbow connected edge-colouring of G is the strong rainbow connection number of G, denoted by src(g) The investigation of src(g) is slightly more challenging than that of rc(g), and fewer papers have been obtained on it For details, see [7, 10, 18, 25] As a natural counterpart of rainbow connection, Krivelevich and Yuster [15] proposed the concept of rainbow vertex-connection A vertex-coloured path is vertex-rainbow if all of its internal vertices have distinct colours A vertex-colouring of G is rainbow vertex-connected if any two vertices of G are connected by a vertex-rainbow path The minimum number of colours in a rainbow vertex-connected vertex-colouring of G is the rainbow vertex-connection number of G, denoted by rvc(g) Corresponding to the strong rainbow connection, Li et al [21] introduced the notion of strong rainbow vertex-connection A vertex-colouring of G is strongly rainbow vertex-connected if for any two vertices u and v of G, there is a vertexrainbow u v geodesic The minimum number of colours in a strongly rainbow vertexconnected vertex-colouring of G is the strong rainbow vertex-connection number of G, denoted by srvc(g) For more results on rainbow vertex-connection, we refer to [22, 26] It was also shown that computing the rainbow connection number and rainbow vertexconnection number of an arbitrary graph is NP-hard [5, 6, 8, 9, 13, 19] For more results on the rainbow connection subject, we refer to the survey [23] and the book [24] Subsequently, Liu et al [27] proposed the concept of total rainbow connection A totalcoloured path is total-rainbow if its edges and internal vertices have distinct colours A total-colouring of G is total rainbow connected if any two vertices of G are connected by a total-rainbow path The minimum number of colours in a total rainbow connected totalcolouring of G is the total rainbow connection number of G, denoted by trc(g) For more results on the total rainbow connection number, see [14, 28] Inspired by the concept of strong rainbow (vertex-)connection, a natural idea is to introduce the strong total rainbow connection number A total-colouring of G is strongly total rainbow connected if for any two vertices u and v of G, there is a total-rainbow u v geodesic The minimum number of colours in a strongly total rainbow connected total-colouring of G is the strong total rainbow connection number of G, denoted by strc(g) Very recently, Dorbec et al [11] initiated the study of rainbow connection in digraphs Subsequently, versions of the other five parameters for digraphs were considered For more details, see [1, 2, 3, 12, 16, 17] This paper will be organised as follows In Section 2, we will present results for all six rainbow connection parameters for general graphs In Section 3, we determine the strong total rainbow connection number of some specific graphs, including cycles, wheels and complete 2

3 bipartite and multipartite graphs Finally in Section 4, we will compare the six parameters, by considering how close and how far apart they can be from one another In particular, we will characterise all pairs of integers a and b such that, there exists a connected graph G with rvc(g) = a and srvc(g) = b, and similarly for the functions trc and strc We mention a few more words on terminology and notation For a graph G, its vertex and edge sets are denoted by V (G) and E(G), and its diameter is denoted by diam(g) Let K n and C n denote the complete graph and cycle of order n (where n 3 for C n ), and K m,n denote the complete bipartite graph with class sizes m and n For two graphs G and H, and a vertex u V (G), we define G u H to be the graph obtained by replacing u with H, and replacing the edges of G at u with all edges between H and the neighbours of u in G We say that G u H is obtained from G by expanding u into H Note that the graph obtained from G by expanding every vertex into H is also known as the lexicographic product G H 2 Remarks and results for general graphs In this section, we present some results about the six rainbow connection parameters rc(g), src(g), rvc(g), srvc(g), trc(g) and strc(g), for general graphs G Let G be a non-trivial connected graph with m edges and n vertices, where q vertices are non-pendent (ie, with degree at least 2) We have the following inequalities diam(g) rc(g) src(g) m, (1) diam(g) 1 rvc(g) srvc(g) min(n 2, q), (2) 2 diam(g) 1 trc(g) strc(g) srvc(g) + m min(m + n 2, m + q), (3) To see the last inequality of (2), the inequality srvc(g) n 2 is a result of Li et al [21] We also have srvc(g) q, since any vertex-colouring of G where all q non-pendent vertices are given distinct colours, is strongly rainbow vertex-connected To see the third inequality of (3), we may take a strongly rainbow vertex-connected colouring of G with srvc(g) colours, and then colour the edges with m further distinct colours This gives a strongly total rainbow connected colouring of G with srvc(g) + m colours The last inequality of (3) then follows from (2) All remaining inequalities are trivial Also, the following upper bound is obvious strc(g) src(g) + q Indeed, a strongly total rainbow connected colouring of G can be obtained from a strongly rainbow colouring with src(g) colours, and then colouring the non-pendent vertices of G with q further distinct colours Similarly, for graphs with diameter 2, we have the following proposition which will be very helpful for later Proposition 1 Let G be a graph with diameter 2 Then strc(g) src(g) + 1 Proof By definition, we may give G a strongly rainbow connected colouring, using src(g) colours Since diam(g) = 2, any two non-adjacent vertices x, y V (G) are connected by a rainbow x y geodesic of length 2 Now, colour all vertices of G with a new colour Then clearly, the resulting total-colouring uses src(g) + 1 colours, and is a strongly total rainbow connected colouring Thus, strc(g) src(g) + 1 3

4 For the functions rc(g) and trc(g), we have the following upper bounds which are better than those of (1) and (3) rc(g) n 1, and trc(g) min(2n 3, n 1 + q) Indeed, we may take a spanning tree T of G, which has n 1 edges and at most min(n 2, q) non-pendent vertices We can assign distinct colours to all edges of T, and to all edges and non-pendent vertices of T, to obtain, respectively, the above two upper bounds As for alternative lower bounds instead of those involving the diameter, we note that for any total rainbow connected colouring of G, the colours of the bridges and cut-vertices must be pairwise distinct Similar observations hold for rainbow connected and rainbow vertexconnected colourings, where respectively, the colours of the bridges, and the colours of the cut-vertices, must be pairwise distinct Hence, the following result holds Proposition 2 Let G be a connected graph Suppose that B is the set of all bridges, and C is the set of all cut-vertices Denote b = B and c = C, respectively Then src(g) rc(g) b, srvc(g) rvc(g) c, strc(g) trc(g) b + c In the next result, we give equivalences and implications when the rainbow connection parameters are small Theorem 3 Let G be a non-trivial connected graph (a) The following are equivalent (i) G is a complete graph (ii) diam(g) = 1 (iii) rc(g) = 1 (iv) src(g) = 1 (v) rvc(g) = 0 (vi) srvc(g) = 0 (vii) trc(g) = 1 (viii) strc(g) = 1 (b) strc(g) trc(g) 3 if and only if G is not a complete graph (c) (i) rc(g) = 2 if and only if src(g) = 2 (ii) rvc(g) = 1, if and only if srvc(g) = 1, if and only if diam(g) = 2 (iii) rvc(g) = 2 if and only if srvc(g) = 2 (iv) trc(g) = 3 if and only if strc(g) = 3 4

5 (v) trc(g) = 4 if and only if strc(g) = 4 Moreover, any of the conditions in (i) implies any of the conditions in (iv), and any of the conditions in (i), (iv) and (v) implies any of the conditions in (ii) Proof Although parts of this result can be found in [7, 21], we provide a proof for the sake of completeness (a) Clearly we have (ii) (i) (iv) Using (1), we can easily obtain (iv) (iii) (ii) Similarly, using (2) and (3), we have (ii) (i) (vi) (v) (ii), and (ii) (i) (viii) (vii) (ii) (b) If G is not a complete graph, then diam(g) 2, and strc(g) trc(g) 3 follows from (3) The converse clearly holds by (a) (c) We first prove (i) Suppose first that src(g) = 2 Then by (a), we have diam(g) 2 By (1), we have 2 rc(g) src(g) = 2, and hence rc(g) = 2 Conversely, suppose that rc(g) = 2 Then (a) and (1) imply that src(g) 2 and diam(g) = 2 Also, there exists a rainbow connected colouring for G, using rc(g) = 2 colours In such an edge-colouring, for any x, y V (G), either xy E(G), or xy E(G) and there is a rainbow x y path of length 2, which is also a rainbow x y geodesic Thus src(g) 2, and src(g) = 2 as required By similar arguments using (a), (2) and (3) we can prove (iv); that the first two conditions of (ii) are equivalent; and that the first condition of (v) implies the second Now we complete the proof of (ii) If rvc(g) = 1, then we can easily use (a) and (2) to obtain diam(g) = 2 If diam(g) = 2, then (2) gives rvc(g) 1 Clearly, the vertexcolouring of G where every vertex is given the same colour is rainbow vertex-connected, and thus rvc(g) 1 Therefore (ii) holds Next, we prove (iii) Suppose first that srvc(g) = 2 Then rvc(g) 2 by (2) Clearly rvc(g) 0 by (a), and rvc(g) 1 by (c)(ii) Thus rvc(g) = 2 Conversely, suppose that rvc(g) = 2 Then by (2), we have srvc(g) 2 and diam(g) 3 We may take a rainbow vertex-connected colouring of G, using at most rvc(g) = 2 colours Let x, y V (G) If d(x, y) {1, 2}, then any x y geodesic is clearly vertex-rainbow If d(x, y) = 3, then since any x y path of length at least 4 cannot be vertex-rainbow, there must exist a vertex-rainbow x y path of length 3, which is also an x y geodesic Thus, the colouring is also strongly rainbow vertex-connected We have srvc(g) 2, so that srvc(g) = 2, and (iii) holds Next, we complete the proof of (v) Suppose that strc(g) = 4 By (a) and (b), we have 3 trc(g) strc(g) = 4 By (iv), we have trc(g) 3, so that trc(g) = 4 Thus (v) holds Finally, we prove the last part of (c) Firstly, suppose that either condition in (i) holds, so that rc(g) = 2 Then (a) and (b) imply trc(g) 3 Moreover, there exists a rainbow connected edge-colouring for G, using rc(g) = 2 colours Clearly by colouring all vertices of G with a third colour, we have a total rainbow connected colouring for G, using 3 colours Thus, trc(g) 3 We have trc(g) = 3, and thus both conditions of (iv) hold Secondly, suppose that any of the conditions in (i), (iv) or (v) holds It is easy to use (a), and (1) or (3), to obtain diam(g) = 2 Thus, the three conditions of (ii) also hold Remark We remark that in Theorem 3(c), no other implication exists between the conditions of (i) to (v) Obviously, no implication exists between the conditions of (ii) and those of (iii) Thus by the last part of (c), no implication exists between the conditions of (iii) and 5

6 those of (i), (iv) and (v) Similarly, no implication exists between the conditions of (iv) and those of (v), and thus no implication exists between the conditions of (i) and those of (v), since the conditions of (i) imply those of (iv) Clearly, the example of the stars K 1,n shows that there are infinitely many graphs where the conditions of (ii) hold, but those of (i), (iv) and (v) do not hold Indeed, for n 2, we have rvc(k 1,n ) = 1, while rc(k 1,n ) = n and trc(k 1,n ) = n + 1 Now, there are infinitely many graphs G such that the conditions of (iv) hold, but those of (i) do not hold For example, let u be a vertex of the cycle C 5, and let G be a graph obtained by expanding u into a clique K That is, G = (C 5 ) u K It was remarked in [27] (and also easy to show) that for any such graph G, we have trc(g) = rc(g) = 3 Now, it is easy to see that if H is a spanning connected subgraph of a connected graph G, then we have rc(g) rc(h), rvc(g) rvc(h), and trc(g) trc(h) However, the following lemma shows that the same inequalities do not hold for the strong rainbow connection parameters Lemma 4 There exist connected graphs G and H such that, H is a spanning subgraph of G, and src(g) > src(h) Similar statements hold for the functions srvc and strc Proof We construct graphs G i and H i, for i = 1, 2, 3, as follows Let H 1 (resp H 2, H 3 ) be the graph as shown in Figure 1(a) (resp (b), (c)) consisting of the solid edges, and G 1 (resp G 2, G 3 ) be the graph obtained by adding the dotted edge (a) x (b) Figure 1 The graphs in Lemma (c) We will prove that src(g 1 ) > src(h 1 ), srvc(g 2 ) > srvc(h 2 ), and strc(g 3 ) > strc(h 3 ) (4) Firstly, it is easy to see that the edge-colouring of H 1 as shown is strongly rainbow connected, and thus src(h 1 ) 4 In fact, we have src(h 1 ) = 4, since src(h 1 ) diam(h 1 ) = 4 Now, suppose that there exists a strongly rainbow connected colouring of G 1, using at most four colours Note that the four pendent edges of G 1 must received distinct colours, say colours 1, 2, 3, 4 The dotted edge has colour 1, 2, 3 or 4, and in each case, we can easily find two vertices that are not connected by a rainbow geodesic in G 1 We have a contradiction, and thus src(g 1 ) 5 > 4 = src(h 1 ) 6

7 We can similarly prove the remaining two inequalities of (4) We have a strongly rainbow vertex-connected colouring of H 2 as shown, and since diam(h 2 ) = 7, we have srvc(h 2 ) = 6 Suppose that there exists a strongly rainbow vertex-connected colouring of G 2, using at most six colours Then, the six cut-vertices of G 2 must received distinct colours, say colours 1, 2, 3, 4, 5, 6 The vertex x has colour 1, 2, 3, 4, 5 or 6, and in each case, we can find two vertices that are not connected by a vertex-rainbow geodesic in G 2 We have a contradiction, and thus srvc(g 2 ) 7 > 6 = srvc(h 2 ) Likewise, we have a strongly total rainbow connected colouring of H 3 as shown, and thus strc(h 3 ) 14 Suppose that there exists a strongly total rainbow connected colouring of G 3, using at most 14 colours Then, the eight bridges and six cut-vertices of G 3 must received distinct colours, say colours 1, 2,, 14 The dotted edge has colour 1, 2,, 13 or 14, and in each case, we can find two vertices that are not connected by a total-rainbow geodesic in G 3 Again we have a contradiction, and thus strc(g 3 ) 15 > 14 strc(h 3 ) Li et al [21] provided a similar example of graphs G and H which gave srvc(g) = 9 > 8 = srvc(h) However in their example, H was not a spanning subgraph of G, although this could be easily corrected Chartrand et al [7] had conjectured that src(g) src(h) whenever G and H are connected graphs, with H a spanning subgraph of G They observed that if this conjecture was true, then we have src(g) n 1 if G is a connected graph of order n However, Lemma 4 shows that the conjecture is false The latter claim may still be true, and we propose this as an open problem, as well as the total-coloured analogue Problem 5 Let G be a connected graph of order n with q non-pendent vertices Then, are the following inequalities true? src(g) n 1, and strc(g) min(2n 3, n 1 + q) 3 Strong total rainbow connection numbers of some graphs In this section, we consider the strong total rainbow connection numbers of some specific graphs, namely, trees, cycles, wheels, and complete bipartite and multipartite graphs The remaining five rainbow connection parameters for these graphs have previously been considered by various authors, and we shall recall these previous results along the way First, let T be a tree of order n, with q non-pendent vertices Note that, since any two vertices of T are connected by a unique path, we have rc(t ) = src(t ), rvc(t ) = srvc(t ), and trc(t ) = strc(t ) From Chartrand et al [7], and Liu et al [26, 27], we have rc(t ) = src(t ) = n 1, rvc(t ) = q, and trc(t ) = n 1 + q Moreover, it is well known that if n 3, then 1 q n 2; and that q = 1 if and only if T is a star, and q = n 2 if and only if T is a path Thus, we have the following result Proposition 6 Let T be a tree with order n, and q non-pendent vertices (a) rvc(t ) = srvc(t ) = q In particular, for n 2, rvc(t ) = srvc(t ) = n 2 if and only if T is a path; and for n 3, rvc(t ) = srvc(t ) = 1 if and only if T is a star (b) trc(t ) = strc(t ) = n 1 + q In particular, for n 2, trc(t ) = strc(t ) = 2n 3 if and only if T is a path; and for n 3, trc(t ) = strc(t ) = n if and only if T is a star 7

8 Our next task is to consider cycles Recall that C n denotes the cycle of order n 3 The functions rc(c n ) and src(c n ) were determined by Chartrand et al [7], while rvc(c n ), srvc(c n ) and trc(c n ) were determined by Li and Liu [20], Lei et al [16], and Liu et al [27], respectively We may summerise these results as follows Theorem 7 [7, 16, 20, 27] (a) rc(c 3 ) = src(c 3 ) = 1, and rc(c n ) = src(c n ) = n 2 for n 4 (b) For 3 n 15, the values of rvc(c n ) and srvc(c n ) are given in the following table n rvc(c n ) srvc(c n ) For n 16, we have rvc(c n ) = srvc(c n ) = n 2 (c) For 3 n 12, the values of trc(c n ) are given in the following table For n 13, we have trc(c n ) = n n trc(c n ) Note that we have the slightly surprising facts that rc(c n ) = src(c n ), but rvc(c n ) = srvc(c n ) except for n = 11, 13, 15; and that srvc(c 11 ) > srvc(c 12 ) By taking advantage of the fact that strc(c n ) trc(c n ) and the proof of part (c) in [27], we have the following result for strc(c n ) Theorem 8 For n 3, we have strc(c n ) = trc(c n ) That is, for 3 n 12, the values of strc(c n ) are given in the table in Theorem 7(c) For n 13, we have strc(c n ) = n Proof One can easily check that strc(c 3 ) = 1, strc(c 4 ) = 3, and strc(c 5 ) = 3 Now, let n 6 We need to prove that strc(c n ) trc(c n ) Thus by Theorem 7(c), we need to prove that strc(c n ) n 1 for 6 n 10 and n = 12, and strc(c n ) n for n = 11 and n 13 The following facts were shown in the proof of Theorem 7(c) in [27] For 6 n 10 and n = 12, there is a total-colouring of C n, using n 1 colours, such that every path of length n 2 1 is total-rainbow, and when n is even, any two opposite vertices of C n are connected by a total-rainbow path For n = 11 and n 13, there is a total-colouring of C n, using n colours, such that every path of length n 2 is total-rainbow With these total-colourings, it is easy to see that any two vertices x and y of C n are connected by a total-rainbow x y path of length at most n 2, which must also be a total-rainbow x y geodesic Thus the total-colourings are also strong total rainbow connected colourings, and the upper bound strc(c n ) trc(c n ) follows 8

9 Next, we consider wheel graphs The wheel W n of order n is the graph obtained from the cycle C n by joining a new vertex v to every vertex of C n The vertex v is the centre of W n Trivially, we have rvc(w 3 ) = srvc(w 3 ) = 0, and rvc(w n ) = srvc(w n ) = 1 for n 4 The functions rc(w n ) and src(w n ) were determined by Chartrand et al [7], while trc(w n ) was determined by Liu et al [27] Theorem 9 [7, 27] (a) rc(w 3 ) = 1, rc(w n ) = 2 for 4 n 6, and rc(w n ) = 3 for n 7 (b) src(w n ) = n 3 for n 3 (c) trc(w 3 ) = 1, trc(w n ) = 3 for 4 n 6, trc(w n ) = 4 for 7 n 9, and trc(w n ) = 5 for n 10 In the next result, we determine the function strc(w n ) The proof is partially based on the fact that strc(w n ) trc(w n ) Theorem 10 strc(w 3 ) = 1, and strc(w n ) = n for n 4 Proof Let v be the centre of W n, and v 1, v 2,, v n be the vertices of W n in the cycle C n Since W 3 is precisely the complete graph K 4, we have strc(w 3 ) = 1 Now, let n 4 Since diam(w n ) = 2, by Proposition 1 and Theorem 9(b), we have strc(w n ) src(w n ) + 1 = n Also, by Theorem 9(c), we have strc(w n) trc(w n ) = 3 = n for 4 n 6 It remains to show that strc(w n) n for n 7 Assume the contrary, and suppose that there is a strongly total rainbow connected colouring c of W n, using at most n 3 colours Since n 7, for each vertex v i, there exists at least one vertex v j with j i such that the unique v i v j geodesic of length 2 passes the centre v Thus, c(v) c(vv i ) for i = 1, 2,, n Therefore, the n edges vv i use at most n 3 1 < n 3 different colours One can deduce that there exist at least four different edges, say vv i, vv j, vv k, vv l, such that c(vv i ) = c(vv j ) = c(vv k ) = c(vv l ) Again, since n 7, we may assume that the unique v i v j geodesic is precisely the path v i vv j So, there is no total-rainbow v i v j geodesic, a contradiction Consequently, strc(w n ) n for n 7 Our next aim is to consider complete bipartite graphs K m,n Clearly we have rc(k 1,n ) = src(k 1,n ) = n; rvc(k 1,1 ) = srvc(k 1,1 ) = 0 and rvc(k m,n ) = srvc(k m,n ) = 1 for (m, n) (1, 1); and trc(k 1,1 ) = strc(k 1,1 ) = 1 and trc(k 1,n ) = strc(k 1,n ) = n + 1 for n 2 For 2 m n, the functions rc(k m,n ) and src(k m,n ) were determined by Chartrand et al [7], and the function trc(k m,n ) was determined by Liu et al [27] Theorem 11 [7, 27] Let 2 m n We have the following (a) rc(k m,n ) = min( m n, 4) (b) src(k m,n ) = m n (c) trc(k m,n ) = min( m n + 1, 7) In the next result, we will determine strc(k m,n ) for 2 m n 9

10 Theorem 12 For 2 m n, we have strc(k m,n ) = m n + 1 Proof Since diam(k m,n ) = 2, we have strc(k m,n ) src(k m,n ) + 1 = m n + 1 by Proposition 1 and Theorem 11(b) Now we prove the lower bound strc(k m,n ) m n + 1 This proof will be a slight modification of the proof of the lower bound of Theorem 11(c) in [27], but we provide it for the sake of clarity Let the classes of K m,n be U = {u 1,, u m } and V, where V = n Let b = m n 2 If m n 2 m, then strc(k m,n ) 3 = b + 1 Now let n > 2 m, so that b 3 We have (b 1) m < n b m Let c be a total-colouring of K m,n, using colours from {1,, b} For v V, assign v with the vector v of length m, where v i = c(u i v) for 1 i m For two partitions P and P of V, we say that P refines P, written P P, if for all A P, we have A B for some B P In other words, P can be obtained from P by partitioning some of the sets of P We define a sequence of refining partitions P 0 P 1 P m of V, with P i (b 1) i for 0 i m, as follows Initially, set P 0 = {V } Now, for 1 i m, suppose that we have defined P i 1 with P i 1 (b 1) i 1 Let P i 1 = {A 1,, A l }, where l (b 1) i 1 Define P i as follows For 1 q l and A q P i 1, let B q 1 = {v A q : v i = c(u i ) or c(u i ) + 1 (mod b)}, B q r = {v A q : v i = c(u i ) + r (mod b)}, for 2 r b 1 Let P i = {Br q : 1 q l, 1 r b 1 and Br q }, so that P i is a partition of V with P i (b 1) i and P i 1 P i Proceeding inductively, we obtain the partitions P 0 P 1 P m of V, with P i (b 1) i for 0 i m Now, observe that for every 1 i m, and any two vertices y and z in the same set in P i, the path yu i z is not total-rainbow, since c(u i y) = y i and c(u i z) = z i are either in {c(u i ), c(u i ) + 1} (mod b), or they are both c(u i )+r (mod b) for some 2 r b 1 Since n > (b 1) m P m, there exists a set in P m with at least two vertices w and x, and since P 1 P m, this means that w and x are in the same set in P i for every 1 i m Therefore, wu i x is not a total-rainbow path for every 1 i m Since the paths wu i x are all the possible w x geodesics (with length 2) in K m,n, it follows that there does not exist a total-rainbow w x geodesic Hence, c is not a strongly total rainbow connected colouring of K m,n, and strc(k m,n ) b + 1 To conclude this section, we consider complete multipartite graphs Let K n1,,n t denote the complete multipartite graph with t 3 classes, where 1 n 1 n t are the class sizes Clearly, we have rvc(k n1,,n t ) = srvc(k n1,,n t ) = 0 (resp 1) if n t = 1 (resp n t 2) The functions rc(k n1,,n t ) and src(k n1,,n t ) were determined by Chartrand et al [7], and the function trc(k n1,,n t ) was determined by Liu et al [27], as follows Theorem 13 [7, 27] Let G = K n1,,n t, where t 3, 1 n 1 n t, m = t 1 i=1 n i and n t = n Then, the functions rc(g), src(g) and trc(g) are given in the following table n = 1 n 2 and m > n m n rc(g) 1 2 min( m n, 3) src(g) 1 2 m n trc(g) 1 3 min( m n + 1, 5) Here, we determine the function strc(k n1,,n t ) for t 3 10

11 Theorem 14 Let t 3, 1 n 1 n t, m = t 1 i=1 n i and n t = n Then, 1 if n = 1, strc(k n1,,n t ) = 3 if n 2 and m > n, m n + 1 if m n Proof Write G for K n1,,n t, and let V i be the ith class (with n i vertices) for 1 i t If n = 1, then G = K t and strc(g) = 1 Now for n 2, we have strc(g) 3 For the case n 2 and m > n, we have src(g) = 2 by Theorem 13 Since diam(g) = 2, by Proposition 1, we have strc(g) src(g) + 1 = 3 Thus, strc(g) = 3 Now, let m n For this case, we have src(g) = m n by Theorem 13 Again by Proposition 1, we have the upper bound strc(g) src(g) + 1 = m n + 1 It remains to prove the lower bound strc(g) m n + 1 Let b = m n 2 If m n 2 m, then strc(g) 3 = b + 1 Now let n > 2 m, so that b 3 We have (b 1) m < n b m Suppose that we have a total-colouring c of G, using at most b colours Note that K m,n is a spanning subgraph of G with classes U = V 1 V t 1 and V t We can restrict the total-colouring c to K m,n and apply the same argument involving the refining partitions as in Theorem 12 We have vertices w, x V t such that all of the paths wux, for u U, are not total-rainbow Since these paths are all the possible w x geodesics in G (of length 2), it follows that there does not exist a total-rainbow w x geodesic in G Therefore, c is not a strongly total rainbow connected colouring of G, and strc(g) b Comparing the rainbow connection numbers Our aim in this section is to compare the various rainbow connection parameters In [15], Krivelevich and Yuster observed that for rc(g) and rvc(g), we cannot generally find an upper bound for one of the parameters in terms of the other Indeed, let s 2 By taking G = K 1,s, we have rc(g) = s and rvc(g) = 1 On the other hand, let the graph G s be constructed as follows Take s vertex-disjoint triangles and, by designating a vertex from each triangle, add a complete graph K s on the designated vertices Then rc(g s ) 4 and rvc(g s ) = s We may consider the analogous situation for the parameters src(g) and srvc(g) Again by taking G = K 1,s, we see that src(g) = s and srvc(g) = 1, so that src(g) can be arbitrarily larger than srvc(g) Rather surprisingly, unlike the situation for the functions rvc(g) and rc(g), we are uncertain if srvc(g) can also be arbitrarily larger than src(g) We propose the following problem Problem 15 Does there exist an infinite family of connected graphs F such that, src(g) is bounded on F, while srvc(g) is unbounded? When considering the total rainbow connection number in addition, we have the following trivial inequalities trc(g) max(rc(g), rvc(g)), (5) strc(g) max(src(g), srvc(g)) (6) In [27], Liu et al considered how close and how far apart the terms in the inequality (5) can be They observed that by considering Krivelevich and Yuster s construction as described 11

12 above, we have trc(g s ) = rvc(g s ) = s for s 13 Also, as mentioned in the remark after the proof of Theorem 3, if G = (C 5 ) u K is a graph obtained by expanding a vertex u of the cycle C 5 into a clique K, then we have trc(g) = rc(g) = 3 Thus, trc(g) can be equal to each of rvc(g) and rc(g) for infinitely many graphs G On the other hand, Liu et al also remarked that, given 1 t < s, there exists a graph G such that trc(g) s and rvc(g) = t Indeed, we can let G = B s,t be the graph obtained by taking the star K 1,s and identifying the centre with one end-vertex of the path of length t (this graph B s,t is a broom) Also, for s 13, we can again consider the graphs G s and obtain trc(g s ) = s and rc(g s ) 4 Thus, trc(g) can also be arbitrarily larger than each of rvc(g) and rc(g) For the difference between the terms trc(g) and max(rc(g), rvc(g)), one can consider G to be the path of length s, and obtain trc(g) = 2s 1 and max(rc(g), rvc(g)) = s, so that trc(g) max(rc(g), rvc(g)) = s 1 can be arbitrarily large However, for this simple example, the term max(rc(g), rvc(g)) is unbounded in s In the final problem in [27], Liu et al asked the question of whether there exists an infinite family of connected graphs F such that, max(rc(g), rvc(g)) is bounded on F, while trc(g) is unbounded This open problem appears to be much more challenging Here, we consider the analogous situations for the terms in the inequality (6) From the previous remarks and results, we can easily obtain the following Theorem 16 (a) There exist infinitely many graphs G with strc(g) = src(g) = 3 (b) Given s 13, there exists a graph G with strc(g) = srvc(g) = s (c) Given 1 t < s, there exists a graph G such that strc(g) s and srvc(g) = t Proof (a) Let G = (C 5 ) u K as described earlier We have trc(g) = rc(g) = 3 By Theorem 3(c), we have strc(g) = 3 Therefore by (1) and (6), we have 3 = strc(g) src(g) rc(g) = 3, so that strc(g) = src(g) = 3 (b) We use the following construction which was given by Lei et al [17] For s 13, let H s be the graph as follows First, we take the graph G s from before, where u 1,, u s are the vertices of the K s, and the remaining vertices are v i, w i, where u i v i w i is a triangle, for 1 i s We then add new vertices z 1,, z s, and connect the edges u i z i, u i+1 z i, v i z i, w i z i+4, for 1 i s, where all indices are taken modulo s In [17], Lei et al proved that strc(h s ) = srvc(h s ) = s (c) Since the broom G = B s,t as described earlier is a tree, it is clear that strc(g) = trc(g) s and srvc(g) = rvc(g) = t As before, if G is the path of length s, then we have strc(g) src(g) = strc(g) max(src(g), srvc(g)) = s 1, so that the two differences can both be arbitrarily large But the terms src(g) and max(src(g), srvc(g)) are unbounded in s Similar to the question of Liu et al in [27] and Problem 15, we may ask the following question Problem 17 Does there exist an infinite family of connected graphs F such that, src(g) is bounded on F, while strc(g) is unbounded? Similarly, does there exist an infinite family of connected graphs F such that, max(src(g), srvc(g)) is bounded on F, while strc(g) is unbounded? 12

13 Now, we proceed to the final part of this section Recall that the following inequalities hold for a connected graph G rc(g) src(g), rvc(g) srvc(g), and trc(g) strc(g) Chartrand et al [7] considered the following question: Given positive integers a b, does there exist a graph G such that rc(g) = a and src(g) = b? They gave positive answers for a = b, and 3 a < b with b 5a 6 3 Chern and Li [10] then improved this result as follows Theorem 18 [10] Let a and b be positive integers Then there exists a connected graph G such that rc(g) = a and src(g) = b if and only if a = b {1, 2} or 3 a b Theorem 18 was an open problem of Chartrand et al, and it completely characterises all possible pairs a and b for the above question Subsequently, Li et al [21] studied the rainbow vertex-connection analogue, and they proved the following result Theorem 19 [21] Let a and b be integers with a 5 and b 7a 8 5 Then there exists a connected graph G such that rvc(g) = a and srvc(g) = b Here, we will improve Theorem 19, and also study the total rainbow connection version of the problem We will prove Theorems 20 and 21 below, where we will completely characterise all pairs of positive integers a and b such that, there exists a graph G with rvc(g) = a and srvc(g) = b (resp trc(g) = a and strc(g) = b) Theorem 20 Let a and b be positive integers Then there exists a connected graph G such that rvc(g) = a and srvc(g) = b if and only if a = b {1, 2} or 3 a b Theorem 21 Let a and b be positive integers Then there exists a connected graph G such that trc(g) = a and strc(g) = b if and only if a = b {1, 3, 4} or 5 a b To prove Theorems 20 and 21, we first prove three auxiliary lemmas Lemma 22 For every b 3, there exists a graph G such that rvc(g) = 3 and srvc(g) = b Proof We construct a graph F b as follows We take a complete graph K 2b, say with vertices u 1,, u 2b, and further vertices v 1,, v 2b, w 1,, w 2b For 1 i 2b, we connect the edges u i v i, u i v i 1, u i w i, w i v i, w i v i 1 Throughout, the indices of the vertices u i, v i, w i are taken modulo 2b We show that rvc(f b ) = 3 and srvc(f b ) = b Suppose firstly that we have a vertex-colouring of F b, using at most two colours Since 2b 6, we may assume that u 1 and u l have the same colour, for some 3 l 2b 1 Then note that w 1 u 1 u l w l is the unique w 1 w l geodesic, with length 3 Thus, there does not exist a vertex-rainbow w 1 w l path, and we have rvc(f b ) 3 Now, we define a vertex-colouring f of F b as follows Let f(u i ) = 1 if i is odd, and f(u i ) = 2 if i is even Let f(z) = 3 for all other vertices z It is easy to check that f is a rainbow vertex-connected colouring for F b For example, to connect w 1 to w i with a vertex-rainbow path, where 3 i 2b 1, we may take w 1 u 1 u i w i if i is even, and w 1 u 1 u i 1 v i 1 w i if i is odd Thus rvc(f b ) 3, and we have rvc(f b ) = 3 Next, suppose that we have a vertex-colouring of F b, using fewer than b colours Then, three of the vertices u i have the same colour, so we may assume that u 1 and u l have the 13

14 same colour, for some 3 l 2b 1 Note that w 1 u 1 u l w l is the unique w 1 w l geodesic (with length 3) Thus, there does not exist a vertex-rainbow w 1 w l geodesic, and we have srvc(f b ) b Now, we define a vertex-colouring g of F b as follows Let g(u i ) = i 2 for 1 i 2b, and g(z) = 1 for all other vertices z We show that g is a strongly rainbow vertexconnected colouring for F b It is easy to see that each vertex u i is at distance at most 2 from every other vertex Thus, it suffices to check that v 1 is connected to each v i and w j by a vertex-rainbow geodesic, and similarly for w 1 to each w j Now, d(v 1, w 1 ) = d(v 1, w 2 ) = 1 and d(v 1, v 2 ) = d(v 1, v 2b ) = 2 Also, d(v 1, v i ) = d(v 1, w j ) = 3 for 3 i 2b 1 and 3 j 2b, whence v 1 u 1 u i v i and v 1 u 1 u j w j are vertex-rainbow v 1 v i and v 1 w j geodesics Likewise, d(w 1, w 2 ) = d(w 1, w 2b ) = 2, and d(w 1, w j ) = 3 for 3 j 2b 1, whence w 1 u 1 u j w j is a vertex-rainbow w 1 w j geodesic Thus srvc(f b ) b, and we have srvc(f b ) = b Lemma 23 For every 4 a b, there exists a connected graph G such that rvc(g) = a and srvc(g) = b Proof We construct a graph F a,b as follows Let n = 2(b 1)(b a + 2) 12 We take a set of vertices V = {v 1,, v n }, and another vertex u and a path u 0 u a 3 We add the paths uw i v i and u a 3 x i v i for 1 i n, and then the edges v l v l+1, w l w l+1, x l x l+1 for 1 l n with l odd Let U = {u 0,, u a 3 }, W = {w 1,, w n } and X = {x 1,, x n } Note that we have perfect matchings within the sets V, W and X We show that rvc(f a,b ) = a and srvc(f a,b ) = b Clearly we have rvc(f a,b ) diam(f a,b ) 1 = a Now, we define a vertex-colouring c of F a,b as follows Let c(u j ) = j for 1 j a 3 For 1 i n, let c(w i+1 ) = c(x i ) = a 2 if i is odd, and c(w i 1 ) = c(x i ) = a 1 if i is even Let c(z) = a for all other vertices z It is easy to check that c is a rainbow vertex-connected colouring for F a,b For example, for i 2, to connect v 1 to v i with a vertex-rainbow path, we may take v 1 x 1 u a 3 x i v i if i is even, and v 1 x 1 u a 3 x i+1 v i+1 v i if i is odd, since a 4 Thus rvc(f a,b ) a, and we have rvc(f a,b ) = a Next, suppose that there exists a strongly rainbow vertex-connected colouring f of F a,b, using at most b 1 colours, say colours 1, 2,, b 1 Then note that for every 1 i n, the unique u 0 v i geodesic is u 0 u 1 u a 3 x i v i Thus we may assume that f(u j ) = j for 1 j a 3, so that f(x i ) {a 2, a 1,, b 1} for 1 i n Also, we have f(w i ), f(u) {1,, b 1} for 1 i n For a 2 p b 1 and 1 q b 2, we define the set A p,q V where A p,1 = {v i V : f(x i ) = p and f(w i ) = f(u) or f(u) + 1 (mod b 1)}, A p,q = {v i V : f(x i ) = p and f(w i ) = f(u) + q (mod b 1)}, for q 2 Note that V = p,q {A p,q : A p,q } is a partition of V with at most (b 2)(b a + 2) parts Since n = 2(b 1)(b a + 2), there exists a set A r,s with at least three vertices Thus, we may assume that v 1, v l A r,s with l 2 Observe that the path v 1 x 1 u a 3 x l v l is not vertex-rainbow, since f(x 1 ) = f(x l ) = r Also, the path v 1 w 1 uw l v l is not vertex-rainbow, since f(w 1 ) and f(w l ) are either in {f(u), f(u) + 1} (mod b 1), or they are both f(u) + s (mod b 1) Since these two paths are the only v 1 v l geodesics (with length 4), we have a contradiction Thus, srvc(f a,b ) b Finally, we define a vertex-colouring g of F a,b, using colours 1, 2,, b, as follows Let g(u j ) = j for 1 j a 3, and g(u) = g(u 0 ) = g(v i ) = b for 1 i n Now, note that 14

15 there are (b 1)(b a + 2) pairs {v l, v l+1 } with l odd, and also (b 1)(b a + 2) distinct vectors of length 2, whose first coordinate is in {a 2,, b 1} and second coordinate is in {1,, b 1} Thus we may assign these distinct vectors to all vertices of V such that, both vertices of a pair {v l, v l+1 } with l odd receive the same vector (so that every vector appears exactly twice) If v l and v l+1 have been assigned with the vector v, then we set g(x l ) = g(x l+1 ) = v 1 {a 2,, b 1}, and g(w l ) = g(w l+1 ) = v 2 {1,, b 1} We show that g is a strongly rainbow vertex-connected colouring for F a,b We must show that for every x, y V (F a,b ), there is a vertex-rainbow x y geodesic If x U and y u, then it is easy to find a vertex-rainbow x y geodesic For example, if x = u j and y = w i, then we take u j u a 3 x i v i w i If x = u j and y = u, then we take u j u a 3 x l v l w l u, where v l is assigned with the vector (a 2, b 1) Similarly, it is easy to deal with the case when x = u and y V W X Now we consider the case x, y V W X Firstly, the cases x, y W and x, y X are clear, since d(x, y) 2 Next, suppose that x V, say x = v 1 Then the case y {w 1, x 1, v 2, w 2, x 2 } is clear, since we have d(x, y) 2 If y = w l (resp x l ) for some l 2, then we take v 1 w 1 uw l (resp v 1 x 1 u a 3 x l ) If y = v l for some l 2, then x and y are assigned with different vectors, say x y If x 1 y 1, then we take v 1 x 1 u a 3 x l v l, and if x 2 y 2, then we take v 1 w 1 uw l v l Finally, it remains to consider the case x W and y X We may assume that x = w 1 and y = x l for some 1 l n We take w 1 v 1 x 1 if l = 1; w 1 v 1 x 1 x 2 if l = 2; and w 1 v 1 x 1 u a 3 x l if l 3 We always have a vertex-rainbow x y geodesic, so that g is a strongly rainbow vertexconnected colouring Therefore srvc(f a,b ) b, and we have srvc(f a,b ) = b Lemma 24 For every 5 a < b, there exists a connected graph G such that trc(g) = a and strc(g) = b Proof We consider the complete multipartite graph K 1,,1,n, where there are m 2 singleton classes, say {u 1 },, {u m } Let U = {u 1,, u m }, and V be the class with n vertices Given 5 a < b, let G a,b,m be the graph constructed as follows We take K 1,,1,n, and set n = (b 2) m + 1 We then add a 1 4 pendent edges at u 1, say W = {w 1,, w a 1 } is the set of pendent vertices We claim that for sufficiently large m, we have trc(g a,b,m ) = a and strc(g a,b,m ) = b Since the bridges of G a,b,m are the a 1 pendent edges, and the only cut-vertex is u 1, clearly we have trc(g a,b,m ) a by Proposition 2 Now we define a total-colouring f of G a,b,m as follows Let f(u 1 w l ) = l for 1 l a 1 For every v V, let f(u 1 v) = 1, and f(u i v) = 2 for all 2 i m Let f(u i u j ) = 4 for all 1 i < j m Let f(u 1 ) = a, and f(z) = 3 for all z V (G a,b,m ) \ {u 1 } We claim that f is a total rainbow connected colouring for G a,b,m We need to show that for every x, y V (G a,b,m ), there is a total-rainbow x y path Since u 1 is connected to all other vertices, it suffices to consider x, y V (G a,b,m ) \ {u 1 } If x, y W and x, y are not adjacent, then x, y V, in which case we take the path xu 1 u 2 y Now suppose x W Then we can take the path xu 1 y, unless if x = w 1 and y V, in which case we take xu 1 u 2 y; or x = w 4 and y U \{u 1 }, in which case we take xu 1 vy for some v V Thus f is a total rainbow connected colouring for G a,b,m, and trc(g a,b,m ) a We have trc(g a,b,m ) = a 15

16 Now, suppose that we have a total-colouring of G a,b,m, using fewer than b colours Note that m n + 1 = b, so that by Theorem 14, for the copy of K 1,,1,n, we have strc(k 1,,1,n ) = b It follows that when restricted to the K 1,,1,n, there are two vertices w, x that are not connected by a total-rainbow w x geodesic This means that we have w, x V, and the paths xuw, for u U, are all not total-rainbow Since these paths are also all the possible w x geodesics in G a,b,m, we do not have a total-rainbow w x geodesic in G a,b,m Thus strc(g a,b,m ) b It remains to prove that strc(g a,b,m ) b Let m be sufficiently large so that (b 1) m 1 > (b 2) m log(b 1) This inequality holds if m > log(b 1) log(b 2) Thus, we have (b 1)m 1 n We define a total-colouring g of G a,b,m as follows Let g(u 1 w l ) = l for 1 l a 1 Let g(u 1 ) = a, and g(u 1 v) = g(u i u j ) = g(z) = b for all v V, 1 i < j m, and z V (G a,b,m ) \ {u 1 } Now since (b 1) m 1 n, we may assign distinct vectors of length m 1 to the vertices of V, with entries from {1, 2,, b 1} Suppose that v V has been assigned with the vector v We let g(u i+1 v) = v i for 1 i m 1 and v V We claim that g is a strongly total rainbow connected colouring for G a,b,m Similar to before, it suffices to show that for all x, y V (G a,b,m ) \ {u 1 }, there is a total-rainbow x y geodesic If x, y W and x, y are not adjacent, then x, y V We have x i y i for some 1 i m 1, so that we can take the geodesic xu i+1 y If x W, then we can take the geodesic xu 1 y Thus g is a strongly total rainbow connected colouring for G a,b,m, and strc(g a,b,m ) b We have strc(g a,b,m ) = b We can now prove Theorems 20 and 21 Proof of Theorem 20 Suppose that there exists a connected graph G such that rvc(g) = a and srvc(g) = b Then obviously we have a b If a = 1 (resp a = 2), then Theorem 3(c)(ii) (resp (c)(iii)) gives b = 1 (resp b = 2) Therefore, we have either a = b {1, 2}, or 3 a b Conversely, given a, b such that either a = b {1, 2} or 3 a b, we show that there exists a connected graph G with rvc(g) = a and srvc(g) = b Obviously if a = b 1, then rvc(g) = srvc(g) = a if G is the path of length a + 1 The remaining cases satisfy 3 a b, and these are covered by Lemmas 22 and 23 Thus Theorem 20 follows Proof of Theorem 21 Suppose that there exists a connected graph G such that trc(g) = a and strc(g) = b Then obviously we have a b If a = 1 (resp a = 3, a = 4), then Theorem 3(a) (resp (c)(iv), (c)(v)) gives b = 1 (resp b = 3, b = 4) Theorem 3(a) and (b) also imply that a, b 2 Therefore, we have either a = b {1, 3, 4}, or 5 a b Conversely, given a, b such that either a = b {1, 3, 4} or 5 a b, we show that there is a connected graph G with trc(g) = a and strc(g) = b Obviously, if a = b = 1, then trc(g) = strc(g) = 1 if G is any non-trivial complete graph, and if a = b 3, then trc(g) = strc(g) = a if G is the star of order a The remaining cases satisfy 5 a < b, and these are covered by Lemma 24 Thus Theorem 21 follows 16

17 Acknowledgements Lin Chen, Xueliang Li and Jinfeng Liu are supported by the National Science Foundation of China (Nos and ) Henry Liu is supported by China Postdoctoral Science Foundation (Nos 2015M580695, 2016T90756), and International Interchange Plan of CSU Henry Liu would also like to thank the Chern Institute of Mathematics, Nankai University, for their generous hospitality He was able to carry out part of this research during his visit there References [1] J Alva-Samos, JJ Montellano-Ballesteros, Rainbow connection in some digraphs, Graphs Combin 32 (2016), [2] J Alva-Samos, JJ Montellano-Ballesteros, A note on the rainbow connectivity of tournaments, arxiv: (2015) [3] J Alva-Samos, JJ Montellano-Ballesteros, Rainbow connection of cacti and some infinity digraphs, submitted [4] B Bollobás, Modern Graph Theory, Springer-Verlag, New York, 1998, xiv+394 pp [5] Y Caro, A Lev, Y Roditty, Zs Tuza, R Yuster, On rainbow connection, Electron J Combin 15 (2008), R57 [6] S Chakraborty, E Fischer, A Matsliah, R Yuster, Hardness and algorithms for rainbow connectivity, 26th International Symposium on Theoretical Aspects of Computer Science STACS 2009 (2009), See also J Combin Optim 21 (2011), [7] G Chartrand, GL Johns, KA McKeon, P Zhang, Rainbow connection in graphs, Math Bohem 133 (2008), [8] L Chen, X Li, H Lian, Further hardness results on the rainbow vertex connection number of graphs, Theoret Comput Sci 481 (2013), [9] L Chen, X Li, Y Shi, The complexity of determining the rainbow vertex-connection of a graph, Theoret Comput Sci 412 (2011), [10] X Chern, X Li, A solution to a conjecture on the rainbow connection number, Ars Combin 104 (2012), [11] P Dorbec, I Schiermeyer, E Sidorowicz, E Sopena, Rainbow connection in oriented graphs, Discrete Appl Math 179 (2014), [12] R Holliday, C Magnant, P Salehi Nowbandegani, Note on rainbow connection in oriented graphs with diameter 2, Theory Appl Graphs 1(1) (2014), article 2 [13] X Huang, X Li, Y Shi, Note on the hardness of rainbow connections for planar and line graphs, Bull Malays Math Sci Soc 38 (2015),

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